In a right triangle ABC, right-angled at B, BC = 12 cm and AB = 5 cm. The radius of the circle inscribed in the triangle (in cm) is

Given Δ ABC, right angled at B, BC = 12 cm and AB = 5 cm.

There is a circle inscribed in the triangle as follows:

Let radius of the circle be r cm.

We know that by Pythagoras Theorem, AC² = AB² + BC²

In Δ ABC,

⇒ AC² = 12² + 5²

⇒ AC² = 144 + 25

⇒ AC² = 169

⇒ AC = 13

From the figure,

Ar (ΔABC) = ar (ΔAOC) + ar (ΔAOB) + ar (ΔBOC)

⇒ 1/2 (AB) (AC) = 1/2 (OP) (AC) + 1/2 (OQ) (BC) + 1/2 (OR) (AB)

⇒ 1/2 (12) (5) = 1/2 (r) (5) + 1/2 (r) (13) + 1/2 (r) (12)

⇒ 1/2 (12) (5) = 1/2 (r) (5 + 13 + 12)

⇒ (12) (5) = r (30)

⇒ r = 2 cm

∴ The radius of the circle inscribed in the given triangle is 2 cm.